Answer
$$\int {{{\left( {\ln x} \right)}^n}} dx = x{\left( {\ln x} \right)^n} - n\int {{{\left( {\ln x} \right)}^{n - 1}}dx} $$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\ln x} \right)}^n}} dx \cr
& {\text{Use integration by parts}} \cr
& u = {\left( {\ln u} \right)^n},{\text{ }}du = n{\left( {\ln x} \right)^{n - 1}}\left( {\frac{1}{x}} \right)dx \cr
& dv = du,{\text{ }}v = u \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{Substituting}} \cr
& \int {{{\left( {\ln x} \right)}^n}} dx = x{\left( {\ln x} \right)^n} - \int {xn{{\left( {\ln x} \right)}^{n - 1}}\left( {\frac{1}{x}} \right)dx} \cr
& \int {{{\left( {\ln x} \right)}^n}} dx = x{\left( {\ln x} \right)^n} - \int {n{{\left( {\ln x} \right)}^{n - 1}}dx} \cr
& \int {{{\left( {\ln x} \right)}^n}} dx = x{\left( {\ln x} \right)^n} - n\int {{{\left( {\ln x} \right)}^{n - 1}}dx} \cr} $$
